Understanding Structural Induction and Strong Induction in Recursively Defined Structures

1. Structural Induction Based on Strong Induction

2. Recursive Definitions Binary tree: 1. Null is an empty binary tree. 2. If Tleftand Trightare binary trees, then <x, Tleft, Tright> is a binary tree 3. The root of <> has value x and left subtree Tleftand right subtree Tright. Arithmetic expression: 1. Any integer n 2. For an expression E, -E is an expression 3. For expressions E, G, E+G, E-G, E*G, E/G (G!=0) are expressions

3. Structural induction is strong induction applied to recursively defined structures. To prove a property P(t) by structural induction we need to prove a base case (or cases) and an induction implication The base cases are derived from the base definition or the simplest instance (like the empty tree or a tree with null subtrees). The induction implication for an instance t generated by n applications of the recursive definition assumes P(s) for any tree s generated by 1 <=k < n applications of the recursive definition. This is strong induction!

4. Example. A node of a tree is any non-null subtree: <x, tl,tr> and we call x the value at that node. A leaf node is a node <x, tl,tr> for which tland trare both null. An internal node is any node that is not a leaf node. We call the number of leaf nodes of a tree T, leaves(T) and the number of internal nodes internals(T).

5. Theorem: For any binary tree T = <x, tl, tr>, leaves(T) <= internals(T) + 1. Call this P(T) Proof by strong induction. Base cases: Consider T0= null (empty tree). leaves(T0) = 0 = internals(T0) and 0 <= 0+1. T1is generated by 1 application of the recursive def , so leaf nodes are null and root node is a leaf. tl= null = tr, then leaves(T1) = 1 and internals(T1) = 0 and 1<=0+1

6. Induction: Assume T is generated by n>= 2 applications of the recursive def. 1. Assume P(S) for any tree, S, generated with k < n applications of the recursive def. 2.2. T = <x, tl, tr > and 2.3 tlgenerated using l applications of def, trusing r applications of def 2.4 T generated by n = l+r+1 applications of rule, def 2.5 l < n, r < n 2.6 leaves(tl)<= internals(tl)+1, leaves(tr)<= internals(tr)+1. 1 and 2.5 2.7 leaves(T) = leaves(tl) + leaves(tr) 2.8 internals(T) = internals(tl) + internals(tr) + 1 since root is not a leaf 2.9 leaves(T) <= internals(tl)+1 + internals(tr)+1. 2.6,2.7 2.10 leaves(T) <= internals(T)+1 ] 2.8, 2.9 For any tree T, P(T). Strong induction since root is not a leaf

7. Structural Induction • The previous example has the idea of structural induction • More formally, we define a well-ordered set S with an order relation, < if for any T ⊆ S, T has a minimal element, e, such that ∀t ∈ T, t!=e => ~(t<e). (Note: there may be more than one minimal element). (In English: e is minimal in T if there is no element in T smaller than e.) • Set of binary trees is well ordered where t1 < t2 means t1 is a subtree of t2. • A minimal element of all trees is the empty tree • Z>=0 is well ordered, where the smaller-than relation is <=. • A minimal element of Z>=0is 0. Minimal element of {2, 4, 6} is 2

8. • Definition 5.6 (Proof by structural induction)(from text) • Suppose that we want to prove that P(x) holds for every x ∈ S, where S is the (well-ordered) set of structures generated by a recursive definition, and P is some property. To give a proof by structural induction of ∀ x ∈ S : P(x), we prove the following: • Base cases: for every x defined by a base case in the definition of S, prove P(x). (base cases are minimal elements) • Inductive cases: for every x defined in terms of y1, y2,…, yk∈ S by an inductive case in the definition of S, prove that P(y1) ∧ P(y2) ∧ ⋯ ∧ P(yk) ⇒ P(x). In other words yi< x for each i.

9. Our previous proof of the statement below was essentially structural induction For any binary tree T = <x, tl, tr>, leaves(T) <= internals(T) + 1. Call this P(T) Proof by structural induction: Base cases are the same: Consider T0= null (empty tree). leaves(T0) = 0 = internals(T0) and 0 <= 0+1. T1is generated by 1 application of the recursive def , so leaf nodes are null and root node is a leaf. tl= null = tr, then leaves(T1) = 1 and internals(T1) = 0 and 1<=0+1 In the proof we would take as the induction assumption P(t) for all t, t a subtree of T. That is any t < T by the ordering over trees. (This gets the explicit use of strong induction out of it.)

10. 1. we make the induction assumption P(t) for all t, t a subtree of T. That is any t < T by the ordering over trees. (This gets the explicit use of strong induction out of it.) 2. T = <x, tl, tr > for some tl, tr 3. leaves(T) = leaves(tl) + leaves(tr) 4. internals(T) = internals(tl) + internals(tr) + 1 5. leaves(tl)<= internals(tl)+1, leaves(tr)<= internals(tr)+1 6. leaves(T) = internals(tl) + 1 + internals(tr) + 1 7. leaves(T) = internals(T) + 1 8. P(t) for all t < T => P(T) (ordering or trees) 9. For any binary tree P(T) by base cases and 8 and 8 and structural induction definition of binary tree since root is not a leaf since root is not a leaf tl, trboth subtrees of T and 1 3, 5 substitution 4, 6 substitution 1 – 7

11. Example: Define a heap as a binary tree <x, tl, tr>, such that for any node in the tree: <y, tyl, tyr>, where tyl= <z, tzl, tzr> or null and tyr= <w, twl, twr> or null then tylis null or y>= z and tyris null or y >= w. Note: this is a local property that simply says the value at a node is greater than or equal the value at the node of either child nodes unless the child is null. Theorem: For any heap H = <x, tl, tr>, x is greater than or equal to the value at any node in the tree. (call this P(H) Prove by structural induction.

12. Base case : <x, tl, tr> is a leaf node (children are null) True since there are no values at the children. Induction case: Assume true for any tree <y, tyl, trr> < H = <x, tl, tr> (That is for any subtree of H) If tl= null, then x >= any value in tl. Otherwise, tl= <y, tyl, trr> and y >= value at every node in tl. Since x >= y by local heap property, x >= value at every node in tl. A similar argument shows that x >= every value in tr. If P(T) for all subtrees T of H, then P(H). By structural induction, P(H) for any heap.